Symmetric Generation and Existence of J3:2, the Automorphism Group of the Third Janko Group

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Journal of Algebra 304 (2006) 256–270 www.elsevier.com/locate/jalgebra

Symmetric generation and existence of J3:2, the automorphism group of the third Janko group

J.D. Bradley, R.T. Curtis ∗

School of Mathematics, University of Birmingham, Edgbaston, Birmingham B15 2TT, UK Received 27 June 2005 Available online 28 February 2006 Communicated by Jan Saxl

Abstract We make use of the primitive permutation action of degree 120 of the automorphism group of L2(16), the special linear group of degree 2 over the field of order 16, to give an elementary definition of the automorphism group of the third Janko group J3 and to prove its existence. This definition enables us to construct the 6156-point graph which is preserved by J3:2, to obtain the order of J3 and to prove its simplicity. A presentation for J3:2 follows from our definition, which also provides a concise notation for the elements of the group. We use this notation to give a representative of each of the conjugacy classes of J3:2. © 2006 Elsevier Inc. All rights reserved.

1. Introduction

The group J3 was predicted by Janko in a paper in 1968 [10]. He considered a ﬁnite simple group with a single class of involutions in which the centraliser of an involution 1+4 was isomorphic to 2 :A5. The following year Higman and McKay, using an observation of Thompson that J3 must contain a subgroup isomorphic to L2(16):2, constructed the group on a computer [9]. Weiss constructed J3 as the automorphism group of a graph with 17442 vertices [12]. This graph has diameter 10 and has a rather complicated structure.

* Corresponding author. E-mail address: [email protected] (R.T. Curtis).

Kleidman and Wilson realised an embedding of J3 into E6(4) [11], which was later proved · without computers by Aschbacher [1]. More recently Baumeister has constructed 3 J3 in terms of an amalgam [2]. ∼ In this paper we give a concise definition of Aut(J3) = J3:2. The minimum degree permutation representation for J3 is 6156, and as complex matrices the minimum degree is 85 (170 for J3:2). Our construction gives a much more concise way of recording elements of J3. · Now J3 possesses a triple cover 3 J3 which has a 9-dimensional irreducible representation over F4, the field of order 4, see the ATLAS [6]. Our definition of Aut(J3) is shown to define either a group of order 100465920 containing a simple subgroup to index 2, or the cyclic group of order 2. The 9-dimensional representation is used to show that the former, in fact, holds. ∼ We consider the (primitive) action of Aut(L2(16)) = L2(16):4 on 120 letters. We take 120 = | 2 = 2 t1,...,t120 ti 1 , the free product of 120 copies of the cyclic group of order 2, 120 and construct a semi-direct product of the form P = 2 :(L2(16):4); as in Curtis [7,8], we refer to such an infinite group as a progenitor. The second author argued that J3:2is a homomorphic image of P as follows. The normaliser of a Sylow 17-subgroup of J3:2 ∼ is of the form 17:8 × 2; the subgroup N = L2(16):4ofJ3:2 contains a maximal subgroup isomorphic to 17:8 which must be contained in this normaliser, and so centralises an invo- 17×16×15×4 = lution. Under conjugation by N this involution thus has 17×8 120 images which, together with the maximal subgroup N, generate J3:2. Indeed, the set of 120 involutions generates a subgroup of J3:2 which is normalised by the whole group and which is not contained in J3; thus it forms a generating set for J3:2. Now elements of P can be written, essentially uniquely, in the form πw where π is a permutation in N and w isawordinthe 120 symmetric generators; thus any relator by which we might factor P has this form. We produce below an easily described relator, found by John Bray, such that

P ∼ G = = J3:2, πwP where πwP denotes the normal closure of the subgroup πw. ∼ The subgroup N = L2(16):4 has index 6156 in J3:2. We are able to construct a graph of rank 7 on 6156 vertices by joining a coset Nw of N to the 120 cosets Nwti , assuming the above relator holds (and, of course, all of its conjugates by permutations of N). This shows that G has order at most 6156 ×|L2(16):4|. It is shown that this graph has diameter 3, so we can represent elements of J3:2 by an element of L2(16):4 followed by a word in the 120 symmetric generators of length at most 3.

2. L2(16):4 and its action on 120 letters

In what follows we shall be deﬁning and constructing a group which contains a set T of ∼ 120 involutions whose set normaliser N is isomorphic to Aut(L2(16)) = L2(16):4 acting primitively on T by conjugation. We shall need to understand this 120 point action in some detail. 258 J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270

Table 1 The permutation character of degree 120 Centraliser 16320 64 60 30 15 17 17 Class 1A 2A 3A 5AB 15ABCD 17ABCD 17EFGH Fixed points 120 8 0 0 0 1 1 Shape 1120 18.256 340 524 158 1.177 1.177

240 16 12 10 24 24 8 8 12 12 2B 4A 6A 10AB 4B 4C 8A 8B 12A 12B 04 00 002 2 0 0 260 14.22.428 620 1012 430 430 12.2.4.814 12.2.4.814 1210 1210

To obtain a copy of the field of order 16 we extend the prime field Z2 by a root, ν say, 4 ∼ 4 of the irreducible quartic x + x + 1. Thus we let F = Z2(ν) = Z2[x]/(x + x + 1).Welet σ be the field automorphism of order 4 which maps each element of F to its square. We ∼ represent each element of N = L2(16):4bya2× 2 matrix of determinant 1 with entries from F , followed by a power of σ . Let the set of Sylow 17-subgroups of N be denoted by S. Then |S|=120 and N acts primitively, by conjugation, on S. In order to obtain a convenient notation for the members of S we proceed as follows. For each H ∈ S we choose the unique element πH ∈ H which, when it acts on the projective line P1(16), maps π2 ∞ to 0. We label H by the ordered pair (0πH , 0 H ). The group N is sharply 3-transitive on P1(16), so this pair defines the subgroup uniquely. The normaliser of one such Sylow 17-subgroup has shape 17:8. Inducing the trivial character of this subgroup up to N (or equivalently, writing down for an element of each conjugacy class of N the number of Sylow 17-subgroups it normalises) we obtain the permutation character shown in the Table 1, where the classes are labelled as in the AT- LAS [6]. This permutation character has norm 4 and using it we may readily write down the cycle shapes of elements in the 120-point action. In particular we see from Table 1 that elements of order 8 have cycle shape 12.2.4.814 and that all fixed point free elements are semi-regular, i.e. their cycles all have the same length. Since elements of order 8 have two fixed points we see that one of the nontrivial orbits of the point stabiliser has length 17, and since the other two have lengths of the form 2a.17 they must contain 34 and 68 points. We say that 2 points are α-joined, β-joined or γ -joined according to whether they lie in the 17-, 34- or 68-orbits of one another’s stabiliser. In Fig. 1 we give a diagram of the resulting α-graph, indicating the orbit lengths of the stabiliser of a point in each of the four suborbits. Thus, for instance the stabiliser of two points x and y which are β-joined to one another has an orbit of length 4 on points α-joined to both x and y, three orbits of length 4 on points γ -joined to x and α-joined to y, and a unique point β-joined to x and α-joined to y. Now an element of order 8 lying in our point stabiliser must, by simple arithmetic, act on these suborbits as 1 + 1 + 82 + 2 + 84 + 4 + 88 and so points in its 2-orbit must be β-joined to its fixed points and points in its 4-orbit must be γ -joined to its fixed points. This shows us that the subgraph on {a,b,c,d,e,f,g,h}, J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 259 1 34 4 43 8 17 1 1 17 8 23 2 68 1 + 24

Fig. 1. The graph on 120 letters.

ac eg

bd fh

Fig. 2. The fix of an involution. the fixed point set of an involution, has form consisting of 4 α-edges ab,cd,ef and gh and otherwise points in the same half are β-joined whilst those in opposite halves are γ -joined, see Fig. 2. An element of order 8 whose 4th power has this fixed point set acts as (a)(b)(c d)(e g f h). By considering the number of involutions and the number of pairs of points each fixes, we can see that every pair of points is fixed by a unique involution. Let us consider how the centraliser of an involution acts on its fixed point set. The normaliser of a Sylow 17-subgroup contains no Klein fourgroup and so commuting involutions cannot fix the same point. Thus the fixed point sets of commuting involutions are disjoint, and the ∼ fixed point sets of the 15 involutions in a Sylow 2-subgroup of N = L2(16) cover the 120 points. In particular every element of the elementary abelian subgroup of order 16 must act trivially or with cycle shape 24 on our fixed point set. So without loss of generality the action of the centraliser of an involution in N on its fixed point set is given by: (a b)(c d)(e f )(g h), (a c)(b d)(e g)(f h), (a e)(b f )(c g)(d h), (c d)(e g f h) .

We now prove a lemma which we shall use in the next section.

Lemma 1. The group N is transitive on ordered triples of the form (a,b,c) where b is α-joined to a and c, and a is γ -joined to c. In particular there is an element which ﬁxes b and interchanges a and c. This element must be an involution.

Proof. The group N is transitive on γ -joined edges. A pair of γ -joined points, (a, c) say, is fixed by a unique involution, π say. From Fig. 1 we can see that there are two points, b and b say, α-joined to both a and c. From Fig. 2 and Table 1 we see that no nontrivial element of N fixes such a configuration, and so the involution fixing a and c must interchange b 260 J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 and b. Hence N is transitive on triples such as a, b, c. Similarly, the element, ρ say, which fixes b and interchanges a and c must be an involution since ρ2 fixes a, b and c.

3. The progenitor and relation

n As in Curtis [7,8] we let m denote a free product of n copies of the cyclic group Cm; that is, a group generated by n elements of order m with no other relation holding between them. In this paper we are only concerned with the case m = 2 and so have n =∼ | 2 = 2 t1,t2,...,tn ti 1 .

A permutation of the n symmetric generators will clearly induce an automorphism on this free product. (In more generality, if m>2 we would consider monomial automorphisms which permute the symmetric generators while raising them to powers co-prime to their order.) If N is a group of permutations of the set {1, 2,...,n} then we may form a group

2n:N,

n π = ∈ a split extension of 2 by N in which ti tiπ for π N. We call the inﬁnite group P a progenitor. We note that every element of P can be written, essentially uniquely, in the form πw, where π ∈ N and w is a word in the symmetric generators. Thus any homomorphic image of P can be written in the form

P = 2n:N . π1w1,π2w2,...

120 From our observations in Section 1 we see that J3:2 is an image of P = 2 :(L2(16):4). We factor this group by a single relation found by John Bray to obtain the group G.We ∼ then ﬁnd an upper bound on the size of G by enumerating the cosets of N = L2(16):4inG. In Section 5 we argue that G must either have that order or be a cyclic group of order 2. In Section 6 we exhibit a group of order greater than 2 with G as an image. Section 4 involves a great deal of calculation involving the group L2(16):4 as a permutation group on 120 letters. Whilst these calculations, if one had the requisite patience, could be done by hand, we use the computer package MAGMA, see [5], to operate within L2(16):4. The relation by which we factor P involves an element of order 12 in L2(16):4. As in Table 1, any element of order 12, μ say, has cycle shape 1210 when acting on 120 points; it turns out that one of the cycles is unique in having the property that if i is a point in 2 the cycle, i is α-joined to iμ and γ -joined to iμ . The relation says that an element, μ,of order 12 in L2(16):4 multiplied by a symmetric generator, ti , in the unique cycle described 5 above has order 5, i.e. (μti) = 1. We note that although elements of order 12 in L2(16):4 are not conjugate to their inverses, using ∼ to mean ‘has the same order as,’ we have −1 −1 −1 μti ∼ (μti) = tiμ ∼ μ ti ; so we may choose μ from either of the conjugacy classes of elements of order 12. J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 261

The element 11 10 of L2(16) has order 3 and clearly commutes with the ﬁeld automorphism σ ,so 11 σ 10 has order 12. It can be checked that the Sylow 17-subgroup we denote by (ν, ν7) lies in the appropriate cycle. Therefore the theorem we seek to prove is:

Theorem 1.

120 2 : (L2(16) : 4) ∼ G = = J3:2. 11 5 10 σt(ν,ν7)

The ﬁrst step of the proof is to ﬁnd an upper bound for the number of cosets of N =∼ L2(16):4inG.

4. Coset enumeration and the graph

∼ A full manual enumeration of the cosets of N = L2(16):4inG can be found in Bradley [3] but, since it is a fairly long and involved calculation, it would not be illu- minating to reproduce it in full here. Instead we describe the general method, prove two lemmas which illustrate this method and list some of the properties of the graph we obtain as a result of the enumeration. In Section 9 we describe how the enumeration was done separately by machine. We note, as in the previous section, that an element of G can be written in the form πw where π ∈ N and w is a word in the symmetric generators, although when we are dealing with an image of P this notation is no longer unique. Hence a coset of N can be written as Nπw = Nw. In the enumeration we consider all double cosets of the form NgN for g ∈ N; by the above these take the simple form NwN. For each double coset NwN,we ﬁnd how many single cosets of N it contains and show that for any symmetric generator ti the coset Nwti is in a double coset we have already considered. We obtain a Cayley graph by joining the coset Nw to the 120 cosets Nwti for ti a symmetric generator. When the graph is closed we have shown that G is ﬁnite and have an upper bound for its order.

Lemma 2. Let ta and tb be two α-joined symmetric generators, and let tc, td , te, tf , tg and th be the other 6 symmetric generators fixed by the unique involution in N which fixes ta and tb in the notation of Fig. 2. Then the double coset NtatbN contains at most 510 single 262 J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 cosets of N and the coset Ntatb is fixed under right multiplication by a subgroup of N of order 32. Furthermore

Ntatb = Ntbta = Ntctd = Ntd tc.

5 Proof. We start with the relation (μti) = 1 which we described above. We multiply this = 5 = μ = μ out to get titj tktltm μ , where tj ti , tk tj , etc. We can rearrange this to find μ7 μ7 = 5 tl tm titj tk μ . As described above, ti and tj are α-joined, as are tj and tk, whereas ti and tk are γ -joined. By Lemma 1 we can find an involution ρ which fixes tj and inter- ρ ρ = ρ 5 ∈ changes ti and tk. If we conjugate by this we find that tktj titl tm (μ ) N. If we multi- μ7 μ7 ρ ρ = 5 ρ 5 ∈ ply this by the rearrangement of our original relation we find tl tm tl tm μ (μ ) N. By construction the first two of these are α-joined, as are the latter two. If we let ta and tb μ7 μ7 denote t and tm , respectively, then ta and tb lie in the fixed point set of a unique involu- l ρ ρ tion, π say. When we calculate tl and tm it turns out that they are tc and td .Wehavethe relation tatbtctd = (a b)(c d), by which we mean an element of order 4 acting as (a b)(c d) on the fixed point set. If we consider an element of order 8 which fixes ta and tb and has 4th power equal to π, then it must preserve the fixed points of π. Then since it fixes no other points than ta and tb, by consideration of the joins in the fixed point set of π it must interchange tc and td . Thus we have Ntatb = Ntbta = Ntctd = Ntd tc. The involution π is centralised by a subgroup of N of order 64. All these elements will preserve the fixed point set of π and half will preserve the set {a,b,c,d}. Hence Ntatb will be stabilised by a subgroup of N of order 32. This means that the double coset NtatbN will contain at most 16,320 = 2 32 510 single cosets of N.

Lemma 3. Let tb and tc be two β-joined symmetric generators, and let ta, td , te, tf , tg and th be the other 6 symmetric generators fixed by the unique involution in N which fixes ta and tb in the notation of Fig. 2. Then the double coset NtbtcN contains at most 2040 single cosets of N and the coset Ntbtc is fixed under right multiplication by a subgroup of N of order 8. Furthermore Ntbtc = Ntbtd .

Proof. From above we have tatbtctd = (a b)(c d). Multiplication by ta gives tbtctd = ta(a b)(c d) = (a b)(c d)tb and so we have tbtctd tb = (a b)(c d) ∈ N or

Ntbtc = Ntbtd .

This coset is ﬁxed under right multiplication by the subgroup of the centraliser of π which 16,320 = commutes with tb, a group of order 8. Hence the double coset NtbtcN will contain 8 2040 single cosets of N. Continuing in this manner we obtain the collapsed Cayley diagram of the graph as shown in Fig. 3. The graph indicates that there are at most 6156 cosets of N inside G and so

|G| 16320 × 6156 = 100,465,920 =|J3:2|. J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 263

The numbers around each double coset represent the orbits of a stabiliser of a single coset on the 120 symmetric generators. We now list some of the properties of the graph which emerge from the enumeration. The cosets in the orbit of length 2720 are of the form Ntate where ta and te are γ -joined symmetric generators. In this case the coset is stabilised by a subgroup of N isomorphic to S3 and there are 3 names for the coset. These names and the S3 which stabilises the coset are not as easy to describe as in the previous two cases, but a description can be found in Bradley [3]. The cosets in the 680 orbit are ﬁxed by one of the 680 subgroups of N of the form 12:2, the normaliser of an element of order 12. The cosets in the orbit of length 85 are of the form Ntatcta where ta and tc are β-joined symmetric generators. If we take the unique involution π which stabilises a and c then CN (π) contains a normal subgroup isomorphic to the Klein fourgroup V4 containing π. The normaliser of this V4 in N is of order 192 and is the stabiliser of the coset Ntatcta. Referring again to Fig. 2, if we take any of the 3 involutions in this V4 ﬁxing x and y say with x and yβ-joined we have Ntatcta = Ntxtytx , giving us 48 names for the coset. 2

5. Existence of J3

Lemma 4. Either G is a group of order 100,465,920 containing a simple subgroup of index 2,orG is the cyclic group of order 2.

Proof. We consider the possibility that a relation which we have not yet discovered holds in G. If any relation holds we can put it in the form πw = 1, where π ∈ N and w is a word in the symmetric generators. From Fig. 3 we know that the length of w is at most 3.

510 16 4 16 4 + + 32 32

8 84 17 + 1 12 120 1 34 2 82 242 1 120 2040 680 12 4 + 1 2 68 87 24

6 + 67 + 3 3 24 3 2 6 96 2720 85 3

34 + 66

Fig. 3. The graph on the cosets of N. 264 J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270

The progenitor P has a normal subgroup consisting of elements of L2(16):2 multiplied by a word of even length in the symmetric generators and elements of L2(16):4 \ L2(16):2 multiplied by a word of odd length in the symmetric generators. The element μti whose fifth power we factor by lies in this normal subgroup, and so any relator πw which is implied must also lie in this subgroup. Firstly we consider the case that a nontrivial element π ∈ L2(16):2 is set equal to the identity. Then the simple group L2(16) must collapse to the identity, and since this group is transitive on the symmetric generators these must all be equal to each other. Then our 5 = 5 = = 5 ∈ : \ : =∼ relation (μti) 1 says that ti ti μ L2(16) 4 L2(16) 2, so we have G C2. The next case to consider is if πta = 1forsomeπ ∈ L2(16):4 \ L2(16):2. Then π ∈ CN (CN (ta)) which is trivial, so some collapse must occur in N and we are in the above ∼ case with G = C2. Thirdly, if πtatb = 1forsomeπ ∈ L2(16):2 then we have Nta = Ntb and since the subgroups of N stabilising these cosets are distinct and maximal we have txty ∈ N ∀x,y. The relation now tells us that titiμ t 2 t 3 t 4 ∈ N and hence ti ∈ N so we are in the second ∼ iμ iμ iμ case and G = C2. Lastly, there is the case that πtatbtc = 1forsomeπ ∈ L2(16):4 \ L2(16):2. From Fig. 3 we see that the coset Ntatbtc = Nta tb tc where a and b are γ -joined, so without loss of generality we may assume that a is γ -joined to b.WehaveNtatb = Ntc, now the stabiliser of Ntatb is of order 6 and the stabiliser of Ntc is maximal and 3 does not divide its order. ∼ So Ntc is stabilised by the whole of N and we are in the above case with G = C2. We have now proved the lemma since we have shown that either we found all the relations which hold in G in our enumeration of the cosets and |G|=100,465,920, or other ∼ relations hold and G = C2. The subgroup of index 2 we described must be simple since if any further relation is imposed on it, it collapses to the trivial group. So if we can find a group, not isomorphic to C2, which is an image of G we have proved the existence of a simple group of order 50,232,960. By the classification of finite simple groups this must be the group J3.

6. A presentation for J3:2

John Bray (unpublished) obtained the following presentation for the automorphism group of the special linear group SL2(16): x,y,z | x17 = y8 = 1,xy = x2,z2 = (xyz)4 = (xz)17 = 1,yz = y5 .

We can check that the following elements of L2(16):4 satisfy the presentation: 0 ν10 ν9 0 10 x = ,y= σ, z = , ν5 ν4 ν14 ν6 ν4 1 where ν is the primitive element of F =∼ GF (16) defined in Section 2, and σ is the field automorphism which squares every element of F . It is readily verified that the subgroup x,y, which is visibly isomorphic to the Frobenius group of order 17 × 8, has index 120; so the presentation does indeed define L2(16):4. In fact x and y generate the normaliser J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 265 7 −3 = 11 of the Sylow 17-subgroup which we denote by (ν, ν ) and zyx 10 σ , the desired element of order 12. This leads to the following presentation for G: G =∼ x,y,z,t | x17 = y8 = 1,xy = x2,z2 = (xyz)4 = (xz)17 = 1,yz = y5, − t2 =[t,x]=[t,y]= zyx 3t 5 = 1 .

By Lemma 4, if we can find elements which satisfy this presentation and generate a ∼ · group bigger than C2, then we have proved that G = J3:2. We use the construction of 3 J3:2 · given in the ATLAS [6, p. 82]. The group 3 J3 is given as a group of 9-dimensional matrices over the field F4 ={0, 1,ω,ω¯ }. We denote the basis of the 9-dimensional vector space by {e1,e2,e3,e4,e5,e6,e7,e0,e∞}. The subspace spanned by ¯ωei + ωei+1 | i ∈ 0, 1, 4, 5 is preserved by a subgroup isomorphic to L2(16):2. We let τ denote the element F1 from · the construction. Then τ is an involution in the outer half of 3 J3:2. Conjugation by τ corresponds to conjugation by the permutation matrix corresponding to the permutation (e1 e3)(e2 e6)(e5 e7) followed by the field automorphism of F4. Consider the following elements: ⎛ ⎞ ωω0 ω¯ ω¯ 00ω¯ 0 ⎜ ⎟ ⎜ ωω¯ 110ω¯ ω¯ ω¯ 0 ⎟ ⎜ ¯ ¯ ⎟ ⎜ ω 1001ω 0 ω 0 ⎟ ⎜ ¯ ⎟ ⎜ ω 000ω 0 ω 00⎟ = ⎜ ¯ ¯ ¯ ¯ ⎟ X ⎜ ω 1 ωωω ω 00ω ⎟ , ⎜ 0 ωωωω¯ 1110⎟ ⎜ ⎟ ⎜ ω¯ 01ω¯ 110ω¯ 1 ⎟ ⎝ ⎠ ω 1 ω¯ 0 ω¯ 00ω 0 0 ω 00000ω¯ 1 ⎛ ⎞ 0 ωωω¯ 1 ω ω¯ 0 ω¯ ⎜ ⎟ ⎜ 0001ω 11ω 0 ⎟ ⎜ ⎟ ⎜ 110ω¯ 1 ω¯ ω¯ 01⎟ ⎜ ⎟ ⎜ 11ω¯ ω¯ ω¯ ωω¯ 00⎟ ⎜ ⎟ Y = ⎜ ω 1 ωωω¯ 10ω¯ 0 ⎟ τ, ⎜ ⎟ ⎜ 0 ω¯ 0 ω 10ω¯ ω¯ 0 ⎟ ⎜ ⎟ ⎜ ωω¯ 1 ω¯ 0 ωω10⎟ ⎝ 01000ω¯ 010⎠ ω¯ 0 ω 0 ω 010ω¯ ⎛ ⎞ 11000ω¯ 000 ⎜ ⎟ ⎜ 11ω¯ 000000⎟ ⎜ ⎟ ⎜ 0 ω 1001000⎟ ⎜ ⎟ ⎜ 000100000⎟ ⎜ ⎟ Z = ⎜ 000000ωωω¯ ⎟ , ⎜ ⎟ ⎜ ω 01001000⎟ ⎜ ⎟ ⎜ 0000ω 00ω¯ ω¯ ⎟ ⎝ 0000ω¯ 0 ω 10⎠ 0000ω¯ 0 ω 01 266 J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 ⎛ ⎞ ω¯ 01ω 0 ω¯ ω¯ ω¯ 1 ⎜ ⎟ ⎜ 0 ω ω¯ ωω¯ ω¯ 01ω ⎟ ⎜ ⎟ ⎜ ωω¯ ω¯ ωω¯ 0000⎟ ⎜ ⎟ ⎜ ω¯ 1 ω ω¯ 0 ω¯ 01ω ⎟ ⎜ ⎟ T = ⎜ 0 ω ω¯ 000ω ω¯ 1 ⎟ τ. ⎜ ⎟ ⎜ ωω010ωω00⎟ ⎜ ⎟ ⎜ ω 0001ω 01ω ⎟ ⎝ 00ω¯ 111ω¯ 11⎠ 001ωωω11ω¯

These satisfy the presentation, apart from the ﬁnal relation. We ﬁnd (ZY X−3T)5 = ωI, where I is the identity matrix. So if we factor by the group of scalar matrices, which has order 3, we have an image of G bigger than C2 and so we have proved Theorem 1. In fact, as in Lemma 4, we have proved—essentially by hand—that the group generated by the displayed matrices (or more properly semi-linear maps) has order 3 × 100,465,920 and contains to index 2 a perfect subgroup with centre of order 3, whose quotient by that centre is simple.

7. Elements of J3:2

The construction of J3:2 given in this paper gives us a very concise way of denoting elements of the group. An element of J3:2 can be written as an element of L2(16):4 followed by a word of length at most 3 in the symmetric generators. Since elements of L2(16):4 can be written as 2×2 matrices followed by a power of the ﬁeld automorphism our notation for elements is considerably shorter than the minimum degree permutation representation of degree 6156, the minimum degree complex matrix representation of degree 170 and even more concise than as 36 × 36 matrices over GF(3). In Bradley [3] a MAGMA [5] program is given for multiplying symmetrically presented elements of J3:2 together. The program also provides a way of switching between symmetrically presented elements of J3:2 and permutations on 6156 letters, so all MAGMA functions relating to permutation groups can be used. In Table 2 we give an element of each conjugacy class of J3:2, demonstrating the con- ciseness of our new notation for elements of the group.

8. Centralisers of involutions

The group J3 was originally deﬁned in terms of the centraliser of an involution. In this ∼ section we give generators for the subgroups of G = J3:2 which centralise involutions occurring naturally in our symmetric presentation. During the enumeration of cosets in tc = tb Bradley [3, p. 52] we ﬁnd that, in the notation of Fig. 2, ta td . We use this fact twice in this section. ∼ ∼ If we take an involution, π say, in N = L2(16) then it is in the simple group G = J3. 1+4 This means the centraliser in J3 is of the form 2 :A5 and the centraliser in J3:2 has form J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 267

Table 2 The conjugacy classes of J3:2 Class Representative Class Representative Class Representative 10 ν5 1 2 1 ν3 2 1A 01 10A σ 8B σ t(ν5,ν7) 10 01 01 ν5 ν5 2 ν 1 2A 10B 5 σ 8C σ 10 ν 1 10 11 11 3A 12At(1,ν)t(1,ν2) 12B σ 10 10 01 ν7 1 ν3 1 3B t(1,ν)t(ν2,ν5) 15A 18A t1,ν2 10 10 10 1 ν3 2 ν14 1 ν3 1 4A σ 15B 18B t1,ν4 01 10 10 ν5 1 ν 1 11 5A 17A 10 18C 10 t(1,ν) 10 ν5 ν5 ν3 1 01 5B 5 17B 24A t(1,ν) ν 1 10 10 11 2 ν 1 01 6A σ 19A σt(1,ν8) 24B t(1,ν3) 10 10 10 ν 1 ν 1 3 8A σt 2 4 19B σt 6 34A ν 1 t 12 5 10 (ν ,ν ) 10 (ν ,1) 10 ν ,ν 11 ν 1 9A σt 5 2Bt(1,ν) 34B t 14 8 10 (ν,ν ) 10 (ν ,ν ) 11 9B σt 4 4Bσ 10 (ν,ν ) 11 2 9Cσt(ν3,1) 6B 10 σ t(1,ν)

1+4 2 :S5. As in Section 2, π will commute with 8 symmetric generators; these elements generate the whole centraliser of π,i.e.

CG(π) =ta,tb,...,th.

2 Consider the involution σ in G . Since all involutions in J3 are conjugate we know 2 ∼ ∼ 1+4 2 CG(σ ) = CG(π) = 2 :S5. The element σ fixes the subfield of F of order 4 so com- ∼ 2 mutes with L = L2(4), a subgroup of N.Nowσ acts fixed point free on the symmetric generators and conjugates each symmetric generator into a symmetric generator to which σ 2 = it is β-joined. Then if ta tc, in the notation of Fig. 2, ta and tc must lie in the fixed point set of a unique involution, which must therefore commute with σ 2; hence its fixed point set 2 σ 2 = σ 2 = = is preserved by σ and so tb td . Then (tatd ta) tctbtc tatd ta by the first paragraph 2 of this section. Hence tatd ta ∈ CG(σ ), and we have 2 = td CG σ L,ta .

We also consider the centraliser of a symmetric generator ta, once more using the notation of Fig. 2. This commutes with a subgroup of G isomorphic to L2(17),so ∼ CG(ta) = L2(17) × 2. Let ρ ∈ N conjugate ta into a symmetric generator β-joined to ρ ta,sota = td say. Now ta and td lie in the ﬁxed point set of a unique involution so, as in ρtbtc = = tc = tb Fig. 2, tb and tc are well deﬁned. Consider ta tctbtd tbtc tctctatctc, since ta td , ρtbtc ∼ ∼ hence ta = ta.NowCN (ta) = 17:8 is maximal in CG(ta) = L2(17) × 2, and so we have CG(ta) = CN (ta), ρtbtc . 268 J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270

9. Mechanical enumeration of the double cosets of N in G

In order to illustrate the manner in which the above enumeration of double cosets can be performed using the double coset enumerator described in Bray and Curtis, see [4], we include the required input and relevant parts of the output. Thus > n:=Group (x*y*z)^4=(x*z)^17=1>; > h:=sub; > Index(n,h); 120 ∼ As in Section 6 the group Aut(L2(16)) = L2(16):4 is deﬁned in terms of three generators x, y and z, such that h =x,y is a Frobenius group of order 17 × 8. The index of 120 is conﬁrmed by the coset enumerator in MAGMA [5]. > f,n120,k:=CosetAction(n,h); > xp:=f(x);yp:=f(y);zp:=f(z); > N:=n120; > zp*yp*xp^-3; (1, 21, 63, 36, 50, 44, 30, 114, 102, 40, 39, 2) (3, 66, 53, 24, 77, 92, 67, 9, 41, 104, 97, 13) (4, 15, 18, 98, 84, 26, 11, 86, 91, 82, 101, 6) (5, 100, 28, 27, 79, 19, 8, 16, 58, 106, 64, 51) (7, 55, 69, 90, 42, 10, 72, 112, 119, 80, 85, 49) (12, 33, 88, 96, 99, 62, 120, 93, 113, 110, 32, 76) (14, 95, 68, 23, 71, 108, 37, 35, 47, 46, 34, 59) (17, 20, 45, 89, 115, 73, 78, 116, 70, 29, 25, 31) (22, 94, 87, 65, 43, 118, 52, 57, 54, 61, 56, 81) (38, 117, 74, 111, 109, 60, 103, 105, 75, 107, 83, 48)

> RR:=[<[1,21,63,36,50],(zp*yp*xp^-3)^5>]; > HH:=[N];

The action of the group n on the cosets of h is returned as n120; the images of x,y and z are defined as xp, yp and zp; and the control subgroup N is defined to be the group n120, of degree 120. As is shown above, the element u = zyx−3 belongs to the class of elements of order 12 referred to in the text. The special 12-cycle is the first cycle displayed; it contains t1.So 5 the relator by which we wish to factor is (ut1) . Note that this relator is equivalent to the relation

5 t1t21t63t36t50 = u , which is the manner in which we input it in the Double Coset Enumerator as RR.The input HH simply gives the subgroup over which we wish to perform the enumeration, in this case the control subgroup N. J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270 269

> CT:=DCEnum(N,RR,HH:Print:=5,Grain:=100);

Index: 6156 === Rank: 7 === Edges: 62

> CT[1]; 6156 > CT[2]; 7 > CT[7]; [ 1, 120, 510, 2720, 680, 2040, 85 ] > CT[4]; [ [], [1], [1,2], [1,26], [ 1, 26, 23 ], [1,6], [1,26,3] ] >

The computer is now instructed to perform an enumeration of the double cosets of N in the group G =N,t1, which by the transitivity of N contains all 120 symmetric generators, subject to the additional relation given in RR. The output CT contains: in CT[1], the index of N in G; in CT[2], the number of suborbits, which is to say the rank of the permutation action of G on the cosets of N in G; in CT[7], lengths of those suborbits; and in CT[4], canonical representatives (CDCRs, see [4]) for these 7 double cosets. Thus, for example, the third double coset contains 510 single cosets and is, in fact, Nt1t2N. Detailed information about the edges of the graph is also returned, but is not included here. It must be stressed that the CDCR is not necessarily the shortest possible, but is essentially the ﬁrst word in the symmetric generators which does not lie in a previously discovered double coset. In this case we see from Fig. 3 that the graph has diameter 3 and that the shortest CDCRs would have lengths 0, 1, 2 (three times) and 3 (twice). So the CDCRs recorded here are indeed the shortest possible. The conventional presentation for the group J3:2 given in Section 6 may be veriﬁed by single coset enumeration using MAGMA [5]. We must adjoin an involution t which commutes with x and y, and then factor by the additional relation. Thus

> j3a:=Group (x*y*z)^4=(x*z)^17=t^2=(x,t)=(y,t)=(z*y*x^-3*t)^5=1>; > nn:=sub; > Index(j3a,nn); 6156 270 J.D. Bradley, R.T. Curtis / Journal of Algebra 304 (2006) 256–270

References

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